Normalized first four wave function for simple harmonic oscillator.

Question:

Write the normalized first four wave function for simple harmonic oscillator and show these wave function graphically. Also state their parity as well as energy.

$$OR$$ Show that normalized harmonic oscillator are orthonormal basis.

\(\Rightarrow\) For ground state n=0, \(E_0(0+\frac12)\hbar\omega=\frac12\hbar\omega\)

$$H_0(\alpha x)=1$$

$$\psi_0(x)=\biggl[\frac{\alpha}{2^0\sqrt{\pi}0!}\biggr]^\frac12 H_0(\alpha x) e^{\frac{-\alpha^2 0^2}{2}}$$

$$or,\;\; \psi_0(x)= \biggl[\frac{\alpha}{\sqrt{\pi}}\biggr]^\frac12 e^{\frac{-\alpha^2 x^2}{2}}$$

$$\therefore\; \psi_0(x)=\biggl[\frac{m\omega}{\hbar\pi}\biggr]^\frac14 e^{\frac{-m\omega}{2\hbar}x^2}\dotsm(2)$$

Even parity state as, \(\hat \pi \psi_0(x)= (+1)\psi_0(x)\)

\(\therefore\) Fr 1st excited state n=1, \(E_1= \frac32 \hbar \omega\)

$$H_1(\alpha x)= 2\alpha x$$

$$\psi_1(x)= \biggl[\frac{\alpha}{2^1 \sqrt{\pi}1!}\biggr]^\frac12 H_1(\alpha x) e^{\frac{-\alpha^2 x^2}{2}}$$

$$=\biggl[\frac{\alpha}{\sqrt{\pi}}\biggr]^\frac12 2^\frac{-1}{2} 2\alpha x e^{-\frac{\alpha^2 x^2}{2}}$$

$$=\biggl[\frac{m\omega}{\hbar \pi}\biggr]^\frac14\sqrt{2}\biggl(\frac{m\omega}{\hbar}\biggr)^\frac12 x e^{\frac{-m\omega}{2\hbar}\cdot x^2}$$

$$=\biggl(\frac{\alpha}{2\sqrt{\pi}}\biggr)^\frac12 2\alpha\cdot \biggl(xe^{-\alpha^2 \frac{x^2}{2}}\biggr)\dotsm(3)$$

1st excited state is odd parity state

For second excited state ( 3rd state):-

Put n=2, \(H_2(\alpha x)= 4\alpha^2 x^2-2\) ; \(E_2= \frac52\hbar \omega\)

$$\psi_2(x)=\biggl[\frac{\alpha}{2^2 \sqrt{\pi}2!}\biggr]^\frac12 H_2(\alpha x)e^{-\frac{\alpha^2 x^2}{2}}$$

$$\therefore\;\; \psi_2(x)= \biggl[\frac{\alpha}{8\sqrt{\pi}}\biggr]^\frac12 (4\alpha^2 x^2-2)e^{-\frac{\alpha^2x^2}{2}}\dotsm(4)$$

$$\therefore\;\; \hat \pi \psi_2(x)= (+1) \psi_2(x): Even\; parity \; state$$

For 3rd excited state (4th state):

Put n=3, \(H_3(\alpha x)= 8\alpha^3 x^3- 12\alpha x \) ; \( E_3= \frac72 \hbar \omega\)

$$\psi_3(x)=\biggl[\frac{\alpha}{2^3 \sqrt{\pi}3!}\biggr]^\frac12 H_3 (\alpha x) e^{-\frac{\alpha^2 x^2 }{2}}$$

$$\therefore \; \psi_3(x)= \biggl[\frac{\alpha}{48\sqrt{\pi}}\biggr]^\frac12 (8\alpha^3 x^3- 12\alpha x) e^{-\frac{\alpha^2 x^2 }{2}}\dotsm(5)$$

It is odd parity state i.e. \(\hat \pi \psi_3(x)= (-1) \psi_3(x)\).

Properties of wave function of S.H.O

1. All energy state represented by even integers \(\psi_0(x), \psi_2(x), \psi_4(x)\dotsm)\) have even parity and those represented by odd integers (\(\psi_1(x), \psi_3(x),\dotsm )\) have odd parity .

2. The number of peaks ( maxima and minima ) on the wave function goes on increasing as the energy of oscillator increases.

For high value of Q. No 'n quantum harmonic oscillator approximate to classical mechanics.

3. Each energy states are non-degenerate ( Each energy state corresponds to different wave function).

4. Different wave function of harmonic oscillator are arthogonal.

5. Each wavefunction for given value of energy cross the classically forbidden region, due to wave nature of oscillator and has finite value of probability of finding outside classical region.

Recursion Formula:

First recursion recursion relation

$$H_{n+1}(x)= 2xH_n(x)- 2n H_{n-1} (x)$$

Second recursion relation

$$H_n'(x)= 2nH_{n-1}(x)$$

$$H_4(x)$$

$$H_{n+1}(x)= 2nH_n(x)-2nH_{n-1}(x)$$

Put n+1=4 \(\Rightarrow \) n= 3

$$H_r(x)= 2xH_3(x)- 2(3) H_2(x)$$

$$=2x(8x^3-12x)-6(4x^2-2)$$

$$= 16x^4 -24x^2 - 24 x^2+ 12$$

$$=16 x^4 - 48 x^2+ 12$$

$$i.e \;\; H_4(\alpha x) = 16 \alpha^4 x^4 - 48\alpha^2 x^2+12$$

Reference:

1. Mathews, P.M and K Venkatesan. A Text Book of Quantum Mechanics. New Delhi: Tata McGraw Hill Publishing Co. Ltd, 1997.
2. Merzbacher, E. Quantum Mechanics . New York: John Wiley, 1969.
3. Prakash, S and S Salauja. Quantum Mechanics. Kedar Nath Ram Nath Publishing Co, 2002.
4. Singh, S.P, M.K and K Singh. Quantum Mechanics. Chand & Company Ltd., 2002.